Explanation 1) Concept: We first find the rate of absorption and rate of emission in terms of photon flux. After taking the ratio of this two quantities, we can find the rate of emission in terms of rate of absorption. Substituting this value in the rate of emission formula, we can find the power of the emitter. 2) Formula: i) E p h = h c λ ii) R e m i t = P e m i t E p h 3) Given: i) Wavelength of sun light λ = 500 n m ii) Area A a b s = 2 × 10 - 6 m 2 iii) Distance of source from detector r = 3 m iv) Rate of absorption R a b s = 4 p h o t o n s / s 4) Calculations: The rate at which photons are absorbed by the detector is related to the rate of photon emission by the light source as follows:. Let I be the photon flux a distance r from the source. Since the photons are emitted uniformly in all directions, we have R e m i t = I A R e m i t = 4 π r 2 I … … … … … … … … … … … … … … … … … … … … … … … … … . 1 ) The detector absorbs the 80 % of flux. R a b s = 0.80 I A a b s … … … … … … … … … … … … … … … … … … … … … … … … … … 2 ) Taking the ratio of 1) and 2) and solving for R e m i t , we get, R a b s = 0.80 A a b s 4 π r 2 R e m i t R e m i t = 4 π r 2 0

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ISBN: 9781119573968

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Publisher: WILEY

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Chapter 38, Problem 7P

To determine

To find:

The power with which the emitter emits light.

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Chapter 38, Problem 6P

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Chapter 38 Solutions

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